Questions: Question 5 1 pts In the following nuclear equation, the unidentified type of radiation represented by a question mark is: ^2813 Al -> ^2814 Si + ? 2^4 He ^0-1 e ^0+1 e gamma ray ^1^n n

Question 5
1 pts

In the following nuclear equation, the unidentified type of radiation represented by a question mark is:

^2813 Al -> ^2814 Si + ?

2^4 He
^0-1 e
^0+1 e
gamma ray
^1^n n

Transcript text: Question 5 1 pts In the following nuclear equation, the unidentified type of radiation represented by a question mark is: \[ { }^{28}{ }_{13} \mathrm{Al} \rightarrow{ }^{28}{ }_{14} \mathrm{Si}+? \] ${ }_{2}^{4} \mathrm{He}$ ${ }^{0}{ }_{-1} e$ ${ }^{0}{ }_{+1} e$ gamma ray ${ }^{1}{ }^{n} n$ Question 6 1 pts

Solution

Solution Steps

Step 1: Analyze the Nuclear Equation

The given nuclear equation is:

\[ {}^{28}_{13} \mathrm{Al} \rightarrow {}^{28}_{14} \mathrm{Si} + ? \]

In this equation, aluminum-28 ($^{28}_{13} \mathrm{Al}$) is transforming into silicon-28 ($^{28}_{14} \mathrm{Si}$).

Step 2: Determine the Type of Radiation

To identify the type of radiation, we need to consider the changes in atomic number and mass number:

The atomic number increases from 13 to 14, indicating the emission of a particle that increases the atomic number by 1.
The mass number remains the same (28), indicating no change in the total number of nucleons.

Step 3: Identify the Radiation Type

The increase in atomic number by 1 with no change in mass number suggests the emission of a positron ($^{0}_{+1} e$), which is a type of beta-plus decay. This process converts a proton into a neutron, increasing the atomic number by 1.